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# Need help lol

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In how many ways can the digits of 45,520 be arranged to form a 5-digit number? (Remember, numbers cannot begin with 0.

Is it 96?

Nov 9, 2018

#1
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That is the answer I find....maybe Rom will find different!

Nov 9, 2018
#2
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4 choices for the 1st digit

4 choices for the 2nd digit

3 choices for the 3rd digit

2 choices for the 4th digit

However because there are 2 5's the number will be halved because of division by 2!

4x4x3x2/2 = 4x4x3 = 48 different 5 digit numbers

Nov 9, 2018
#3
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hmm, I have one shot and I'm trying to figure this one out.  Rom, you say to divide by 2! but why do you divide by 2! when you don't use the second 5 at all?

Nov 9, 2018
#4
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Cool!   You DO use the second '5' ...but it is idistiguishable from the other....so the numbers would look the same....so you must divide by two....

ElectricPavlov  Nov 9, 2018
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to expand on this a bit permutations of the same digits are indistinguishable.

5,5 is just 5,5, same with 5,5,5, or 5,5,5,5.  They all have only a single permutation.

Whereas n different digits would have n! permutations.

When you go through and take the product of the possibilities like I did it assumes all permutations

are distinguishable.  That's why you then have to divide the result by n!, in this case 2! as there are 2 5's.

If there were 3 5's I'd divide by 3! etc.

Rom  Nov 9, 2018
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Nov 9, 2018
#6
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I would go with that !

ElectricPavlov  Nov 9, 2018
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+279
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Thanks, that was definetly it.  I really appreciate the help!

Nov 9, 2018